Vertical asymptotes

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We explore functions that “shoot to infinity” near certain points.

Video Lecture

Consider the function

$f(x) = \frac {1}{(x+1)^2}.$

While the $\lim _{x\rightarrow -1} f(x)$ does not exist, we learned in the last section that we can still say something about the behavior of the function.

Which of the following are correct?

On the other hand, consider the function

$f(x) = \frac {1}{(x-1)}.$

While the two sides of the limit as $x$ approaches $1$ do not agree, we can still consider the one-sided limits. We see $\lim _{x\rightarrow 1^+} f(x) = \infty$ and $\lim _{x\rightarrow 1^-} f(x) = -\infty$ .

If at least one of the following hold:

$\lim _{x\rightarrow a} f(x) = \pm \infty$ ,
$\lim _{x\rightarrow a^+} f(x) = \pm \infty$ ,
$\lim _{x\rightarrow a^-} f(x) = \pm \infty$ ,

then the line $x=a$ is a vertical asymptote of $f$ .

Find the vertical asymptotes of

$f(x) = \frac {x^2-9x+14}{x^2-5x+6}.$

Since $f$ is a rational function, it is continuous on its domain. So the only points where the function can possibly have a vertical asymptote are zeros of the denominator.
Start by factoring both the numerator and the denominator:

$\frac {x^2-9x+14}{x^2-5x+6} = \frac {(x-2)(x-7)}{(x-2)(x-3)}$ Using limits, we must investigate what happens with $f(x)$ when $x\rightarrow 2$ and $x\rightarrow 3$ , since $2$ and $3$ are the only zeros of the denominator. Write

$\begin{align*} \lim_{x\rightarrow 2} \frac{\cancel{(x-2)}(x-7)}{\cancel{(x-2)}(x-3)} &= \lim_{x\rightarrow 2} \frac{(x-7)}{(x-3)}\\ &=\answer{5}. \end{align*}$

Now write

$\begin{align*} \lim_{x\rightarrow 3} \frac{\cancel{(x-2)}(x-7)}{\cancel{(x-2)}(x-3)} &= \lim_{x\rightarrow 3} \answer{\frac{(x-7)}{(x-3)}} \end{align*}$

Consider the one-sided limits separately.

When $x\rightarrow 3^+$ , the quantity $(x-3)$ is positive and approaches $0$ and the numerator is negative, therefore, $\lim _{x\rightarrow 3^+} f(x) = -\infty$ .

On the other hand, when $x\rightarrow 3^-$ , the quantity $(x-3)$ is negative and approaches $0$ and the numerator is negative, therefore, $\lim _{x\rightarrow 3^-} f(x) = \infty$ .

Hence we have a vertical asymptote at $x=3$ .

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